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3 years ago

15

Unlike most real bulbs, the resistances of the bulbs in the questions below do not change as the current through them changes. A

ll bulbs considered in this problem are identical. Assume all batteries are ideal (they have no internal resistance) and connecting wires have no resistance.
1. The current passing through the brighter

bulb is larger.

2. Not enough information is given.

3. The current passing through the brighter

bulb is smaller.

4. The current passing through both bulbs

are the same.

1 answer:

Shkiper50 [21]3 years ago

5 0

Answer:

Option 1 is correct.

The current passing through the brighter bulb is larger.

Explanation:

The brightness of the bulb is determined by the power, I²R

And since they all have equal resistances, the only factor different that could result in more or less power is the current, I through the bulb.

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Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of

mart [117]

Answer:

The value is $B = 3.33 *10^{-6} \ T$

Explanation:

From the question we are told that

The distance of separation is $d = 0.6 \ m$

The current on the one wire is $I_1 = 9 \ A$

The current on the second wire is $I_2 = 4 \ A$

Generally the magnitude of the field exerted between the current carrying wire is

$B = B_1 - B_2$

Here $B_1$ is the magnetic field due to the first wire which is mathematically represented as

$B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}$

Here $d_1$ is the distance to the half way point of the separation and the value is

$d_1 = 0.3 \ m$

$B_2$ is the magnetic field due to the first wire which is mathematically represented as

$B_2 = \frac{\mu_o * I_2 }{2 \pi * d_2}$

Here $d_2$ is the distance to the half way point of the separation and the value is

$d_2 = 0.3 \ m$

This means that $d_1 = d_2 = a = 0.3$

So

$B = \frac{\mu_o * I_1 }{2 \pi * d_1} - \frac{\mu_o * I_2 }{2 \pi * d_2}$

=> $B = \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }$

=> $B = \frac{ 4\pi * 10^{-7} * (9- 4)}{2 * 3.142 *0.3 }$

=> $B = 3.33 *10^{-6} \ T$

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3 years ago

A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc

guajiro [1.7K]

Answer:

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

$a_o=a\times e^{-kt}$

where,

k = rate constant

t = age of sample

$a_o$ = let initial amount of the reactant

a = amount left after decay process

We have :

$a_o=x$

$a=58\%\times x=0.58x$

t = 95 s

$0.58x=x\times e^{-k\times 95 s}$

$\k= 0.005734 s^{-1}$

Half life is given by for first order kinetics::

$t_{1/2}=\frac{0.693}{k}$

$=\frac{0.693}{0.005734 s^{-1}}=120.86 s$

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

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